Question

tan^3A/1+tan^2A+cot^3A/1+cot^2A=1-2sin^2Acos^2A/sinAcosA​

Answer 1

Answer:

L.H.S=tan^3A/(1+tan^2A)+cot^3A/(1+cot^2A) =tan^3A/sec^2A+cot^3A/cosec^2A =(sin^3A/cos^3A)/(1/cos^2A)+(cos^3A/sin^A)/(1/sin^2A) =sin^3A/cosA+cos^3A/sinA =(sin^4A+cos^A)/sinA.cosA =[(sin^2A+cos^2A)^2-2sin^2A.cos^2A]/sinA.cosA =(1-2sin^A.cos^A)/sinA.cosA R.H.S=secA.cosecA-2sinA.cosA =1/cosA.1/sinA-2sinAcosA =(1-sin^2A.cos^2A)/sinA.cosA Hence,L.H.S=R.H.S.......

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Answer 2

Answer:

your answer attached in the photo